Question

what is the irrational cube root ^3 radical -16 rewritten as a product of an integer and another irrational cube root

Answer 1

To rewrite \(\sqrt[3]{-16}\) as a product of an integer and another irrational cube root, we can first factor \(-16\) into its components.

We note that:
\[
-16 = -1 \cdot 16
\]

Next, we can express \(16\) as \(2^4\):
\[
-16 = -1 \cdot 2^4
\]

Now we can rewrite the cube root:
\[
\sqrt[3]{-16} = \sqrt[3]{-1 \cdot 2^4}
\]

Using the property of cube roots, this can be simplified to:
\[
\sqrt[3]{-16} = \sqrt[3]{-1} \cdot \sqrt[3]{2^4}
\]

Calculating each part:
1. \(\sqrt[3]{-1} = -1\)
2. \(\sqrt[3]{2^4} = \sqrt[3]{2^3 \cdot 2} = \sqrt[3]{2^3} \cdot \sqrt[3]{2} = 2 \cdot \sqrt[3]{2}\)

Now, substituting these values back in:
\[
\sqrt[3]{-16} = -1 \cdot (2 \cdot \sqrt[3]{2}) = -2 \cdot \sqrt[3]{2}
\]

Therefore, we can rewrite \(\sqrt[3]{-16}\) as:
\[
\sqrt[3]{-16} = -2 \sqrt[3]{2}
\]

Thus, \(\sqrt[3]{-16}\) rewritten as a product of an integer and another irrational cube root is:
\[
-2 \sqrt[3]{2}
\]